Advertisements
Advertisements
प्रश्न
Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.
उत्तर
Let P(n): `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24` ∀ n ∈ N
Step 1: P(2) : `1/(2 + 1) + 1/(2 + 2) > 13/24`
⇒ `1/3 + 1/4 > 13/24`
⇒ `7/12 > 13/24`
⇒ `14/24 > 13/24` which is true for P(2).
Step 2: P(k) : `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) > 13/24`.
Let it be true for P(k).
Step 3: P(k + 1) : `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) + 1/(2(k + 1)) > 13/24`
Since `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) > 13/24`
So `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) + 1/(2(k + 1)) > 13/24`
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.
APPEARS IN
संबंधित प्रश्न
Prove the following by using the principle of mathematical induction for all n ∈ N:
1.2 + 2.3 + 3.4+ ... + n(n+1) = `[(n(n+1)(n+2))/3]`
Prove the following by using the principle of mathematical induction for all n ∈ N:
Prove the following by using the principle of mathematical induction for all n ∈ N: `1+2+ 3+...+n<1/8(2n +1)^2`
If P (n) is the statement "n(n + 1) is even", then what is P(3)?
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.
2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.
Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N?
Prove by method of induction, for all n ∈ N:
12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
Prove by method of induction, for all n ∈ N:
1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`
Prove by method of induction, for all n ∈ N:
1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`
Prove by method of induction, for all n ∈ N:
(24n−1) is divisible by 15
Prove by method of induction, for all n ∈ N:
5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`
Answer the following:
Prove, by method of induction, for all n ∈ N
2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer
Prove the statement by using the Principle of Mathematical Induction:
n3 – 7n + 3 is divisible by 3, for all natural numbers n.
A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.
Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin ((n + 1))/2 theta)/(sin theta/2)`, for all n ∈ N.
Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.